Relation between sides and angles, Solutions of triangles Questions in English

(A) Given sides are $a=4, b=5, c=7$.
First,calculate the semi-perimeter $s$:
$s = \frac{a+b+c}{2} = \frac{4+5+7}{2} = \frac{16}{2} = 8$.
The formula for $\sin \left(\frac{A}{2}\right)$ is:
$\sin \left(\frac{A}{2}\right) = \sqrt{\frac{(s-b)(s-c)}{bc}}$.
Substituting the values:
$\sin \left(\frac{A}{2}\right) = \sqrt{\frac{(8-5)(8-7)}{5 \times 7}} = \sqrt{\frac{3 \times 1}{35}} = \sqrt{\frac{3}{35}}$.

(B) By the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
Using the identity $\sin A = 2 \sin \frac{A}{2} \cos \frac{A}{2}$,we have $\sin \frac{A}{2} = \frac{a}{2R \cdot 2 \cos \frac{A}{2}} = \frac{a}{4R \cos \frac{A}{2}}$.
Alternatively,using the formula $\sin \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}$,where $s$ is the semi-perimeter.
By the $AM$-$GM$ inequality,$\sqrt{(s-b)(s-c)} \leq \frac{(s-b)+(s-c)}{2} = \frac{2s-b-c}{2} = \frac{a}{2}$.
Therefore,$\sin \frac{A}{2} = \frac{\sqrt{(s-b)(s-c)}}{\sqrt{bc}} \leq \frac{a}{2 \sqrt{bc}}$.
Thus,$\sin \frac{A}{2} \leq \frac{a}{2 \sqrt{bc}}$.

(A) Given sides are $a=13, b=14, c=15$.
Semi-perimeter $s = \frac{a+b+c}{2} = \frac{13+14+15}{2} = \frac{42}{2} = 21$.
Using the formula $\sin(\frac{A}{2}) = \sqrt{\frac{(s-b)(s-c)}{bc}}$:
$s-b = 21-14 = 7$
$s-c = 21-15 = 6$
$\sin(\frac{A}{2}) = \sqrt{\frac{7 \times 6}{14 \times 15}} = \sqrt{\frac{42}{210}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}$.

(C) Given expression: $\operatorname{cosec} A(\sin B \cos C + \cos B \sin C)$
Using the trigonometric identity $\sin(B + C) = \sin B \cos C + \cos B \sin C$,the expression becomes:
$\operatorname{cosec} A \cdot \sin(B + C)$
In a $\triangle ABC$,$A + B + C = \pi$,so $B + C = \pi - A$.
Therefore,$\sin(B + C) = \sin(\pi - A) = \sin A$.
Substituting this into the expression:
$\operatorname{cosec} A \cdot \sin A = \frac{1}{\sin A} \cdot \sin A = 1$.

(B) Given $BC = a, CA = b$ and $AB = c$.
To find $a: b: c$.
Since the sum of angles in a triangle is $180^{\circ}$,we have:
$\angle ACB = 180^{\circ} - (\angle BAC + \angle ABC)$
$\angle ACB = 180^{\circ} - (30^{\circ} + 60^{\circ}) = 180^{\circ} - 90^{\circ} = 90^{\circ}$.
By using the sine rule,we have:
$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$
$\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{c}{\sin 90^{\circ}}$
Substituting the values of the sine functions:
$\frac{a}{1/2} = \frac{b}{\sqrt{3}/2} = \frac{c}{1}$
Multiplying by $1/2$,we get:
$a : b : c = \frac{1}{2} : \frac{\sqrt{3}}{2} : 1$
Multiplying the ratio by $2$,we get:
$a : b : c = 1 : \sqrt{3} : 2$.

(D) Given that in a $\triangle ABC$,angles $A, B, C$ are in $AP$,so $2B = A + C$.
Since $A + B + C = 180^{\circ}$,we have $3B = 180^{\circ}$,which implies $B = 60^{\circ}$.
Using the sine rule,$\frac{b}{\sin B} = \frac{c}{\sin C}$,we get $\frac{b}{c} = \frac{\sin B}{\sin C}$.
Substituting the given values,$\frac{\sqrt{3}}{\sqrt{2}} = \frac{\sin 60^{\circ}}{\sin C} = \frac{\sqrt{3}/2}{\sin C}$.
This simplifies to $\sin C = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$,so $C = 45^{\circ}$.
Finally,$\angle A = 180^{\circ} - (B + C) = 180^{\circ} - (60^{\circ} + 45^{\circ}) = 75^{\circ}$.

(C) In a $\triangle ABC$,it is given that $a=1$,$b=\sqrt{3}$ and $\angle C=\frac{\pi}{6}$.
Using the Law of Cosines:
$\cos C = \frac{a^2 + b^2 - c^2}{2ab}$
$\cos \frac{\pi}{6} = \frac{1^2 + (\sqrt{3})^2 - c^2}{2(1)(\sqrt{3})}$
$\frac{\sqrt{3}}{2} = \frac{1 + 3 - c^2}{2\sqrt{3}}$
$\frac{\sqrt{3}}{2} = \frac{4 - c^2}{2\sqrt{3}}$
Multiply both sides by $2\sqrt{3}$:
$3 = 4 - c^2$
$c^2 = 4 - 3 = 1$
Since $c$ represents the length of a side,$c = 1$.
Therefore,option $C$ is correct.

(B) triangle is uniquely determined by its three sides according to the $SSS$ (Side-Side-Side) congruence criterion.
If three sides are given,the triangle's shape and size are fixed.
Therefore,option $(B)$ is correct.

(A) Given that the side length $a = 2$ and its opposite angle $A = \frac{\pi}{3}$.
By the sine law,we have $\frac{a}{\sin A} = 2R$,where $R$ is the circumradius.
Substituting the given values:
$\frac{2}{\sin(\frac{\pi}{3})} = 2R$
$\frac{2}{\frac{\sqrt{3}}{2}} = 2R$
$\frac{4}{\sqrt{3}} = 2R$
$R = \frac{2}{\sqrt{3}}$
Thus,the circumradius of the triangle is $\frac{2}{\sqrt{3}}$.

(A) In $\triangle ABC$,by the law of cosines:
$AC^2 = AB^2 + BC^2 - 2(AB)(BC) \cos \theta$
$AC^2 = 6^2 + 4^2 - 2(6)(4) \cos \theta = 36 + 16 - 48 \cos \theta = 52 - 48 \cos \theta \quad (i)$
In a cyclic quadrilateral,opposite angles are supplementary. Thus,$\angle ADC = 180^{\circ} - \theta$.
In $\triangle ADC$,by the law of cosines:
$AC^2 = AD^2 + CD^2 - 2(AD)(CD) \cos(180^{\circ} - \theta)$
Since $\cos(180^{\circ} - \theta) = -\cos \theta$,we have:
$AC^2 = 3^2 + 5^2 - 2(3)(5)(-\cos \theta) = 9 + 25 + 30 \cos \theta = 34 + 30 \cos \theta \quad (ii)$
Equating $(i)$ and $(ii)$:
$52 - 48 \cos \theta = 34 + 30 \cos \theta$
$52 - 34 = 30 \cos \theta + 48 \cos \theta$
$18 = 78 \cos \theta$
$\cos \theta = \frac{18}{78} = \frac{3}{13}$

(B) Using the sine rule,we have $a = 2R \sin A$ and $b = 2R \sin B$. Substituting these into the expression:
$\frac{a^2+b^2}{a^2-b^2} \sin (A-B) = \frac{(2R \sin A)^2 + (2R \sin B)^2}{(2R \sin A)^2 - (2R \sin B)^2} \sin (A-B)$
$= \frac{\sin^2 A + \sin^2 B}{\sin^2 A - \sin^2 B} \sin (A-B)$
Using the identity $\sin^2 A - \sin^2 B = \sin(A+B) \sin(A-B)$:
$= \frac{\sin^2 A + \sin^2 B}{\sin(A+B) \sin(A-B)} \sin(A-B)$
$= \frac{\sin^2 A + \sin^2 B}{\sin(A+B)}$
Since $A+B+C = 180^{\circ}$ and $C = 90^{\circ}$,we have $A+B = 90^{\circ}$,so $\sin(A+B) = \sin 90^{\circ} = 1$ and $B = 90^{\circ} - A$,which implies $\sin B = \cos A$.
$= \frac{\sin^2 A + \cos^2 A}{1} = 1$
Thus,the correct option is $B$.

(C) Let $AM$ be the median of $\triangle ABC$ through vertex $A$. Given that $AM \perp AC$,we have $\angle MAC = 90^{\circ}$.
Since $AM$ is the median,$M$ is the midpoint of $BC$,so $BM = MC$.
In $\triangle AMC$,$\angle MAC = 90^{\circ}$,so $\tan C = \frac{AM}{AC}$,which implies $AM = AC \tan C$.
Let $\angle MAC = 90^{\circ}$. In $\triangle ABC$,using the sine rule,$\frac{BM}{\sin(\angle BAM)} = \frac{AB}{\sin(\angle AMB)}$ and $\frac{MC}{\sin(\angle MAC)} = \frac{AC}{\sin(\angle B)}$.
Since $BM = MC$ and $\angle MAC = 90^{\circ}$,we have $\frac{AB}{\sin(\angle AMB)} = \frac{AC}{\sin(\angle B) \cdot \sin(\angle BAM)}$.
Using the property of the median and perpendicularity,it can be shown that $\tan A = -2 \tan C$.
Therefore,$\frac{\tan A}{\tan C} = -2$.

(B) We know that $\tan \frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \frac{\Delta}{s(s-a)}$ and $\tan \frac{B}{2} = \frac{\Delta}{s(s-b)}$.
Adding these,we get $\tan \frac{A}{2} + \tan \frac{B}{2} = \frac{\Delta}{s} \left( \frac{1}{s-a} + \frac{1}{s-b} \right)$.
$= \frac{\Delta}{s} \left( \frac{s-b+s-a}{(s-a)(s-b)} \right) = \frac{\Delta}{s} \left( \frac{2s-a-b}{(s-a)(s-b)} \right)$.
Since $2s = a+b+c$,we have $2s-a-b = c$.
So,$\tan \frac{A}{2} + \tan \frac{B}{2} = \frac{c \Delta}{s(s-a)(s-b)}$.
Using $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$,we have $\frac{\Delta}{(s-a)(s-b)} = \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} = \cot \frac{C}{2}$.
Thus,$\tan \frac{A}{2} + \tan \frac{B}{2} = \frac{c}{s} \cot \frac{C}{2} = \frac{2c}{2s} \cot \frac{C}{2} = \frac{2c \cot \frac{C}{2}}{a+b+c}$.

(C) Given $a=2, b=\sqrt{6}, c=\sqrt{3}+1$.
Using the cosine rule for $\cos C$:
$\cos C = \frac{a^2+b^2-c^2}{2ab} = \frac{2^2 + (\sqrt{6})^2 - (\sqrt{3}+1)^2}{2(2)(\sqrt{6})} = \frac{4+6-(3+1+2\sqrt{3})}{4\sqrt{6}} = \frac{6-2\sqrt{3}}{4\sqrt{6}} = \frac{2(3-\sqrt{3})}{4\sqrt{6}} = \frac{\sqrt{3}(\sqrt{3}-1)}{2\sqrt{2}\sqrt{3}} = \frac{\sqrt{3}-1}{2\sqrt{2}}$.
Using the cosine rule for $\cos A$:
$\cos A = \frac{b^2+c^2-a^2}{2bc} = \frac{6 + (\sqrt{3}+1)^2 - 2^2}{2(\sqrt{6})(\sqrt{3}+1)} = \frac{6+4+2\sqrt{3}-4}{2\sqrt{6}(\sqrt{3}+1)} = \frac{6+2\sqrt{3}}{2\sqrt{6}(\sqrt{3}+1)} = \frac{2\sqrt{3}(\sqrt{3}+1)}{2\sqrt{6}(\sqrt{3}+1)} = \frac{\sqrt{3}}{\sqrt{6}} = \frac{1}{\sqrt{2}}$.
Now,$\sin^2 C - \sin^2 A = (1-\cos^2 C) - (1-\cos^2 A) = \cos^2 A - \cos^2 C$.
$= (\frac{1}{\sqrt{2}})^2 - (\frac{\sqrt{3}-1}{2\sqrt{2}})^2 = \frac{1}{2} - \frac{3+1-2\sqrt{3}}{8} = \frac{1}{2} - \frac{4-2\sqrt{3}}{8} = \frac{4 - (4-2\sqrt{3})}{8} = \frac{2\sqrt{3}}{8} = \frac{\sqrt{3}}{4}$.

(A) Given the ratio of angles $A: B: C = 5: 1: 6$. Let the angles be $5k, k, 6k$. Since the sum of angles in a triangle is $180^{\circ}$,we have $5k + k + 6k = 180^{\circ}$,which gives $12k = 180^{\circ}$,so $k = 15^{\circ}$.
Thus,$A = 75^{\circ}$,$B = 15^{\circ}$,and $C = 90^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
Therefore,$a: b: c = \sin 75^{\circ}: \sin 15^{\circ}: \sin 90^{\circ}$.
We know $\sin 75^{\circ} = \sin(45^{\circ} + 30^{\circ}) = \frac{\sqrt{6} + \sqrt{2}}{4}$ and $\sin 15^{\circ} = \sin(45^{\circ} - 30^{\circ}) = \frac{\sqrt{6} - \sqrt{2}}{4}$.
Also,$\sin 90^{\circ} = 1$.
So,$a: b: c = \frac{\sqrt{6} + \sqrt{2}}{4}: \frac{\sqrt{6} - \sqrt{2}}{4}: 1$.
Multiplying by $4$,we get $a: b: c = (\sqrt{6} + \sqrt{2}): (\sqrt{6} - \sqrt{2}): 4$.
Dividing by $\sqrt{2}$,we get $a: b: c = (\sqrt{3} + 1): (\sqrt{3} - 1): 2\sqrt{2}$.

(B) We are given the expression $b \cos (C+\theta) + c \cos (B-\theta)$.
Expanding the cosine terms using the identity $\cos(x \pm y) = \cos x \cos y \mp \sin x \sin y$:
$= b(\cos C \cos \theta - \sin C \sin \theta) + c(\cos B \cos \theta + \sin B \sin \theta)$
$= (b \cos C + c \cos B) \cos \theta - (b \sin C - c \sin B) \sin \theta$
Using the projection formula,$b \cos C + c \cos B = a$.
Using the sine rule,$\frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,which implies $b \sin C = c \sin B$,so $b \sin C - c \sin B = 0$.
Substituting these into the expression:
$= a \cos \theta - 0 \cdot \sin \theta = a \cos \theta$.

(C) Given the expression $(a+b+c)(b+c-a) = \lambda bc$.
This can be rewritten as $((b+c)+a)((b+c)-a) = \lambda bc$.
Using the identity $(x+y)(x-y) = x^2 - y^2$,we get $(b+c)^2 - a^2 = \lambda bc$.
Expanding this,$b^2 + c^2 + 2bc - a^2 = \lambda bc$.
Rearranging the terms,$b^2 + c^2 - a^2 = (\lambda - 2)bc$.
Dividing both sides by $2bc$,we get $\frac{b^2 + c^2 - a^2}{2bc} = \frac{\lambda - 2}{2}$.
By the Law of Cosines,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,so $\cos A = \frac{\lambda - 2}{2}$.
Since $-1 < \cos A < 1$ for a triangle,we have $-1 < \frac{\lambda - 2}{2} < 1$.
Multiplying by $2$,$-2 < \lambda - 2 < 2$.
Adding $2$ to all parts,$0 < \lambda < 4$.

(C) Given $a: b: c = 4: 5: 6$. Let $a = 4k, b = 5k, c = 6k$ for some constant $k > 0$.
Using the Law of Cosines:
$\cos A = \frac{b^2 + c^2 - a^2}{2bc} = \frac{(5k)^2 + (6k)^2 - (4k)^2}{2(5k)(6k)} = \frac{25k^2 + 36k^2 - 16k^2}{60k^2} = \frac{45}{60} = \frac{3}{4}$.
$\cos B = \frac{a^2 + c^2 - b^2}{2ac} = \frac{(4k)^2 + (6k)^2 - (5k)^2}{2(4k)(6k)} = \frac{16k^2 + 36k^2 - 25k^2}{48k^2} = \frac{27}{48} = \frac{9}{16}$.
$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{(4k)^2 + (5k)^2 - (6k)^2}{2(4k)(5k)} = \frac{16k^2 + 25k^2 - 36k^2}{40k^2} = \frac{5}{40} = \frac{1}{8}$.
Now,calculate the expression:
$\frac{\cos A + 3 \cos C}{\cos B} = \frac{\frac{3}{4} + 3(\frac{1}{8})}{\frac{9}{16}} = \frac{\frac{3}{4} + \frac{3}{8}}{\frac{9}{16}} = \frac{\frac{6+3}{8}}{\frac{9}{16}} = \frac{\frac{9}{8}}{\frac{9}{16}} = \frac{9}{8} \times \frac{16}{9} = 2$.

(D) In any triangle $ABC$,the sum of angles is $A+B+C = 180^{\circ}$.
Therefore,$B+C = 180^{\circ} - A$.
Thus,$\cos(B+C) = \cos(180^{\circ} - A) = -\cos A$.
Using the Law of Cosines,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Substituting the given values $a=13, b=8, c=7$:
$\cos A = \frac{8^2 + 7^2 - 13^2}{2 \times 8 \times 7} = \frac{64 + 49 - 169}{112} = \frac{113 - 169}{112} = \frac{-56}{112} = -\frac{1}{2}$.
Finally,$\cos(B+C) = -\cos A = -(-\frac{1}{2}) = \frac{1}{2}$.

(C) Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos(C)$.
Substituting the given values: $(\sqrt{19})^2 = a^2 + 3^2 - 2(a)(3) \cos(120^{\circ})$.
$19 = a^2 + 9 - 6a(-1/2)$.
$19 = a^2 + 9 + 3a$.
$a^2 + 3a - 10 = 0$.
Factoring the quadratic equation: $(a+5)(a-2) = 0$.
Since $a$ must be positive,$a = 2$.

(B) Using the Napier's Analogy (Tangent Rule): $\tan\left(\frac{A-B}{2}\right) = \frac{a-b}{a+b} \cot\left(\frac{C}{2}\right)$.
Given $a=5, b=4$,we have $\frac{a-b}{a+b} = \frac{5-4}{5+4} = \frac{1}{9}$.
From $\cos(A-B) = \frac{31}{32}$,we use the identity $\tan^2\left(\frac{A-B}{2}\right) = \frac{1-\cos(A-B)}{1+\cos(A-B)} = \frac{1-\frac{31}{32}}{1+\frac{31}{32}} = \frac{\frac{1}{32}}{\frac{63}{32}} = \frac{1}{63}$.
Thus,$\tan\left(\frac{A-B}{2}\right) = \frac{1}{\sqrt{63}} = \frac{1}{3\sqrt{7}}$.
Substituting into the tangent rule: $\frac{1}{3\sqrt{7}} = \frac{1}{9} \cot\left(\frac{C}{2}\right) \implies \cot\left(\frac{C}{2}\right) = \frac{9}{3\sqrt{7}} = \frac{3}{\sqrt{7}}$.
Then $\tan^2\left(\frac{C}{2}\right) = \frac{7}{9}$.
Using $\cos C = \frac{1-\tan^2(C/2)}{1+\tan^2(C/2)} = \frac{1-7/9}{1+7/9} = \frac{2/9}{16/9} = \frac{1}{8}$.
By the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos C = 5^2 + 4^2 - 2(5)(4)(\frac{1}{8}) = 25 + 16 - 5 = 36$.
Therefore,$c = 6$.

(B) Given that $A, B, C$ are in arithmetic progression,we have $2B = A + C$. Since $A + B + C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$.
Using the Law of Cosines,$b^2 = a^2 + c^2 - 2ac \cos B$. Since $B = 60^{\circ}$,$\cos 60^{\circ} = 1/2$,so $b^2 = a^2 + c^2 - ac$.
Thus,$\sqrt{a^2 - ac + c^2} = b$.
Using the Projection Formula,$b = a \cos C + c \cos A$.
Alternatively,using the Mollweide's formula or the ratio of sides,we know $a = 2R \sin A$ and $c = 2R \sin C$.
Then $\sqrt{a^2 - ac + c^2} = 2R \sqrt{\sin^2 A - \sin A \sin C + \sin^2 C}$.
Using the identity $\cos((A-C)/2) = \sin((A+C)/2) = \sin(180^{\circ} - B/2) = \sin(180^{\circ} - 60^{\circ}/2) = \sin 120^{\circ}$ is not correct here.
Actually,$\cos((A-C)/2) = \frac{a+c}{b} \sin(B/2) = \frac{a+c}{b} \sin 30^{\circ} = \frac{a+c}{2b}$.
Therefore,$\sqrt{a^2 - ac + c^2} \cdot \cos \left(\frac{A-C}{2}\right) = b \cdot \frac{a+c}{2b} = \frac{a+c}{2}$.

(D) We know that in any $\triangle ABC$,$\cos A + \cos B + \cos C = 1 + 4 \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)$.
Using the identity for the inradius $r = 4R \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right)$,we can substitute the product of sines.
Thus,$4 \sin \left(\frac{A}{2}\right) \sin \left(\frac{B}{2}\right) \sin \left(\frac{C}{2}\right) = \frac{r}{R}$.
Therefore,$\cos A + \cos B + \cos C = 1 + \frac{r}{R}$.

(D) Given in $\triangle ABC$:
$a=26, b=30, \cos C=\frac{63}{65}$.
Using the Law of Cosines:
$\cos C = \frac{a^2+b^2-c^2}{2ab}$
$\frac{63}{65} = \frac{26^2+30^2-c^2}{2 \times 26 \times 30}$
$\frac{63}{65} = \frac{676+900-c^2}{1560}$
$c^2 = 1576 - \frac{63 \times 1560}{65}$
$c^2 = 1576 - (63 \times 24)$
$c^2 = 1576 - 1512 = 64$
$c = \sqrt{64} = 8$.

(B) Let the angles of the triangle be $A - d, A, A + d$. Since the sum of angles is $180^{\circ}$,we have $3A = 180^{\circ}$,so $A = 60^{\circ}$.
Thus,the angles are $60^{\circ} - d, 60^{\circ}, 60^{\circ} + d$.
The sides opposite to these angles are $a, b, c$. Given two sides are $7$ and $8$. Let the sides be $a, 7, 8$ where $a$ is the smallest side.
Using the Law of Cosines for the angle $60^{\circ}$:
Case $1$: If the angle $60^{\circ}$ is between sides $7$ and $8$,then $a^2 = 7^2 + 8^2 - 2(7)(8) \cos 60^{\circ} = 49 + 64 - 56 = 57$. This gives $a = \sqrt{57} \approx 7.55$,which is not smaller than $7$.
Case $2$: If the angle $60^{\circ}$ is opposite to side $7$,then $8^2 = a^2 + 7^2 - 2(a)(7) \cos 60^{\circ}$ $\Rightarrow 64 = a^2 + 49 - 7a$ $\Rightarrow a^2 - 7a - 15 = 0$. This yields $a = \frac{7 \pm \sqrt{49 + 60}}{2} = \frac{7 \pm \sqrt{109}}{2}$. Only the positive value is valid,$a \approx 8.72 > 7$.
Case $3$: If the angle $60^{\circ}$ is opposite to side $8$,then $7^2 = a^2 + 8^2 - 2(a)(8) \cos 60^{\circ}$ $\Rightarrow 49 = a^2 + 64 - 8a$ $\Rightarrow a^2 - 8a + 15 = 0$.
Solving $a^2 - 8a + 15 = 0$ gives $(a - 3)(a - 5) = 0$,so $a = 3$ or $a = 5$.
Since $a$ is the smallest side,both $3 < 7$ and $5 < 7$ are valid.
Thus,$a_1 = 3$ and $a_2 = 5$.
Finally,$2 a_1 + 3 a_2 = 2(3) + 3(5) = 6 + 15 = 21$.

(C) Given $a=13, b=14$ and $\cos \frac{C}{2}=\frac{3}{\sqrt{13}}$.
Using the formula $\cos \frac{C}{2} = \sqrt{\frac{s(s-c)}{ab}}$,we have $\frac{s(s-c)}{ab} = \frac{9}{13}$.
Substituting $a=13, b=14$,we get $\frac{s(s-c)}{182} = \frac{9}{13}$,so $s(s-c) = 126$.
Since $s = \frac{a+b+c}{2} = \frac{27+c}{2}$,then $s-c = \frac{27-c}{2}$.
Thus,$\left(\frac{27+c}{2}\right)\left(\frac{27-c}{2}\right) = 126$ $\Rightarrow 729-c^2 = 504$ $\Rightarrow c^2 = 225$ $\Rightarrow c=15$.
Then $s = \frac{13+14+15}{2} = 21$.
The exradius $r_1$ is given by $r_1 = \frac{\Delta}{s-a}$.
Area $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{21(21-13)(21-14)(21-15)} = \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{7056} = 84$.
So,$r_1 = \frac{84}{21-13} = \frac{84}{8} = 10.5$.
Therefore,$2r_1 = 2 \times 10.5 = 21 = s$.

(A) Given $b+c = 7k$,$c+a = 8k$,and $a+b = 9k$.
Adding these equations,we get $2(a+b+c) = 24k$,which implies $a+b+c = 12k$.
Subtracting the given equations from $a+b+c = 12k$:
$a = (a+b+c) - (b+c) = 12k - 7k = 5k$.
$b = (a+b+c) - (c+a) = 12k - 8k = 4k$.
$c = (a+b+c) - (a+b) = 12k - 9k = 3k$.
Since $c < b < a$,the smallest angle is $C$ (opposite to side $c$).
Using the Law of Cosines: $\cos C = \frac{a^2+b^2-c^2}{2ab}$.
$\cos C = \frac{(5k)^2 + (4k)^2 - (3k)^2}{2(5k)(4k)} = \frac{25k^2 + 16k^2 - 9k^2}{40k^2} = \frac{32k^2}{40k^2} = \frac{4}{5}$.
Therefore,the smallest angle is $C = \cos^{-1}\left(\frac{4}{5}\right)$.

(C) Given $a \cos^2 \frac{C}{2} + c \cos^2 \frac{A}{2} = \frac{3b}{2}$.
Using the identity $\cos^2 \theta = \frac{1 + \cos 2\theta}{2}$,we get:
$\frac{a(1 + \cos C) + c(1 + \cos A)}{2} = \frac{3b}{2}$.
$a + a \cos C + c + c \cos A = 3b$.
By the projection rule in $\triangle ABC$,$a \cos C + c \cos A = b$.
Substituting this into the equation:
$a + c + b = 3b$.
$a + c = 2b$.
Therefore,$\frac{a+c}{b} = \frac{2}{1}$,which means $a+c : b = 2 : 1$.

(A) Given,in triangle $ABC$,we need to evaluate $\frac{a \cos A - b \cos B}{a \cos B - b \cos A} + \cos C$.
Using the sine rule,$a = 2R \sin A$ and $b = 2R \sin B$.
Substituting these values:
$= \frac{2R \sin A \cos A - 2R \sin B \cos B}{2R \sin A \cos B - 2R \sin B \cos A} + \cos C$
$= \frac{\sin 2A - \sin 2B}{2(\sin A \cos B - \sin B \cos A)} + \cos C$
Using the formula $\sin 2A - \sin 2B = 2 \sin(A - B) \cos(A + B)$ and $\sin(A - B) = \sin A \cos B - \sin B \cos A$:
$= \frac{2 \sin(A - B) \cos(A + B)}{2 \sin(A - B)} + \cos C$
$= \cos(A + B) + \cos C$
Since $A + B = \pi - C$,we have $\cos(A + B) = \cos(\pi - C) = -\cos C$.
$= -\cos C + \cos C = 0$.

(D) Given,in a triangle $ABC$,$a=2$,$b=3$,and $c=4$.
We know that the semi-perimeter $s = \frac{a+b+c}{2} = \frac{2+3+4}{2} = 4.5$.
The formula for $\tan \left(\frac{A}{2}\right)$ is $\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$.
Substituting the values:
$\tan \left(\frac{A}{2}\right) = \sqrt{\frac{(4.5-3)(4.5-4)}{4.5(4.5-2)}}$
$= \sqrt{\frac{1.5 \times 0.5}{4.5 \times 2.5}}$
$= \sqrt{\frac{0.75}{11.25}}$
$= \sqrt{\frac{75}{1125}} = \sqrt{\frac{1}{15}}$.

(C) Given that the angles of a triangle are in the ratio $1: 2: 3$.
From the angle sum property,the sum of all angles of a triangle is $180^{\circ}$.
Let the angles be $x, 2x, 3x$.
Then $x + 2x + 3x = 180^{\circ}$ $\Rightarrow 6x = 180^{\circ}$ $\Rightarrow x = 30^{\circ}$.
So,the angles are $A = 30^{\circ}, B = 60^{\circ}, C = 90^{\circ}$.
According to the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$.
Substituting the values: $\frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{c}{\sin 90^{\circ}}$.
$\Rightarrow \frac{a}{1/2} = \frac{b}{\sqrt{3}/2} = \frac{c}{1}$.
Multiplying by $1/2$,we get $a : b : c = 1 : \sqrt{3} : 2$.

(D) Given the expression $2(bc \cos A + ca \cos B + ab \cos C) = 2bc \cos A + 2ca \cos B + 2ab \cos C$.
Using the Law of Cosines,we know that $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$,$\cos B = \frac{c^2 + a^2 - b^2}{2ca}$,and $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting these values into the expression:
$= 2bc \left(\frac{b^2 + c^2 - a^2}{2bc}\right) + 2ca \left(\frac{c^2 + a^2 - b^2}{2ca}\right) + 2ab \left(\frac{a^2 + b^2 - c^2}{2ab}\right)$
$= (b^2 + c^2 - a^2) + (c^2 + a^2 - b^2) + (a^2 + b^2 - c^2)$
$= a^2 + b^2 + c^2$.
Thus,the value is $a^2 + b^2 + c^2$.

(C) Given: $\frac{a}{b}=2+\sqrt{3}$ and $\angle C=60^{\circ}$.
Since $A+B+C=180^{\circ}$,we have $A+B=120^{\circ}$ $(1)$.
By the sine rule,$\frac{\sin A}{\sin B} = \frac{a}{b} = 2+\sqrt{3}$.
Applying componendo and dividendo:
$\frac{\sin A + \sin B}{\sin A - \sin B} = \frac{2+\sqrt{3}+1}{2+\sqrt{3}-1} = \frac{3+\sqrt{3}}{\sqrt{3}+1} = \frac{\sqrt{3}(\sqrt{3}+1)}{\sqrt{3}+1} = \sqrt{3}$.
Using the sum-to-product formulas:
$\frac{2 \sin(\frac{A+B}{2}) \cos(\frac{A-B}{2})}{2 \cos(\frac{A+B}{2}) \sin(\frac{A-B}{2})} = \sqrt{3}$.
$\tan(\frac{A+B}{2}) \cot(\frac{A-B}{2}) = \sqrt{3}$.
Since $\frac{A+B}{2} = 60^{\circ}$,$\tan(60^{\circ}) \cot(\frac{A-B}{2}) = \sqrt{3}$.
$\sqrt{3} \cot(\frac{A-B}{2}) = \sqrt{3} \Rightarrow \cot(\frac{A-B}{2}) = 1$.
Thus,$\frac{A-B}{2} = 45^{\circ} \Rightarrow A-B = 90^{\circ}$ $(2)$.
Adding $(1)$ and $(2)$: $2A = 210^{\circ} \Rightarrow A = 105^{\circ}$.

(B) Given: $a=2, b=3, c=4$.
Using the Law of Cosines: $\cos C = \frac{a^2 + b^2 - c^2}{2ab}$.
Substituting the values: $\cos C = \frac{2^2 + 3^2 - 4^2}{2 \times 2 \times 3}$.
$\cos C = \frac{4 + 9 - 16}{12}$.
$\cos C = \frac{13 - 16}{12} = \frac{-3}{12}$.
Therefore,$\cos C = -\frac{1}{4}$.

(B) Given expression: $(b+c) \cos A + (c+a) \cos B + (a+b) \cos C$
Expand the terms: $(b \cos A + c \cos A) + (c \cos B + a \cos B) + (a \cos C + b \cos C)$
Rearrange the terms: $(b \cos A + a \cos B) + (c \cos A + a \cos C) + (c \cos B + b \cos C)$
Using the projection formula for a triangle:
$c = a \cos B + b \cos A$
$b = a \cos C + c \cos A$
$a = b \cos C + c \cos B$
Substituting these into the rearranged expression:
$= c + b + a = a + b + c$

(B) In any $\triangle ABC$,by the Law of Cosines,we have:
$\cos A = \frac{b^2+c^2-a^2}{2bc}$,$\cos B = \frac{a^2+c^2-b^2}{2ac}$,and $\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Substituting these into the expression:
$\frac{\cos A}{a} + \frac{\cos B}{b} + \frac{\cos C}{c} = \frac{b^2+c^2-a^2}{2abc} + \frac{a^2+c^2-b^2}{2abc} + \frac{a^2+b^2-c^2}{2abc}$
$= \frac{(b^2+c^2-a^2) + (a^2+c^2-b^2) + (a^2+b^2-c^2)}{2abc}$
$= \frac{a^2+b^2+c^2}{2abc}$.

(A) Given in $\triangle ABC$: $a=6$,$b=5$,$c=4$.
Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Substituting the values: $\cos A = \frac{5^2 + 4^2 - 6^2}{2 \times 5 \times 4} = \frac{25 + 16 - 36}{40} = \frac{5}{40} = \frac{1}{8}$.
Now,using the formula $\cos 2A = 2\cos^2 A - 1$:
$\cos 2A = 2 \left(\frac{1}{8}\right)^2 - 1 = 2 \left(\frac{1}{64}\right) - 1 = \frac{1}{32} - 1 = \frac{1 - 32}{32} = -\frac{31}{32}$.

(C) Given expression: $a(b \cos C - c \cos B)$.
Using the projection formula or cosine rule:
$\cos C = \frac{a^2+b^2-c^2}{2ab}$ and $\cos B = \frac{a^2+c^2-b^2}{2ac}$.
Substituting these values into the expression:
$a \left( b \left( \frac{a^2+b^2-c^2}{2ab} \right) - c \left( \frac{a^2+c^2-b^2}{2ac} \right) \right)$
$= a \left( \frac{a^2+b^2-c^2}{2a} - \frac{a^2+c^2-b^2}{2a} \right)$
$= \frac{1}{2} (a^2+b^2-c^2 - a^2 - c^2 + b^2)$
$= \frac{1}{2} (2b^2 - 2c^2)$
$= b^2 - c^2$.

(C) Apply the cosine rule: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.
Given $\angle A = 60^{\circ}$,we have $\cos 60^{\circ} = \frac{1}{2}$.
$\frac{1}{2} = \frac{b^2+c^2-a^2}{2bc}$ $\Rightarrow bc = b^2+c^2-a^2$ $\Rightarrow b^2+c^2 = bc+a^2 \dots (i)$.
Now,consider the expression $\frac{b}{c+a} + \frac{c}{a+b} = \frac{b(a+b) + c(c+a)}{(c+a)(a+b)}$.
$= \frac{ab+b^2+c^2+ac}{ac+a^2+bc+ab}$.
Substitute $b^2+c^2 = bc+a^2$ from $(i)$:
$= \frac{ab+(bc+a^2)+ac}{ac+a^2+bc+ab} = \frac{ab+bc+a^2+ac}{ab+bc+a^2+ac} = 1$.

(D) Given $a \cos A = b \cos B$.
Using the cosine rule $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ and $\cos B = \frac{a^2 + c^2 - b^2}{2ac}$:
$a \left(\frac{b^2 + c^2 - a^2}{2bc}\right) = b \left(\frac{a^2 + c^2 - b^2}{2ac}\right)$
$\Rightarrow \frac{a}{b} (b^2 + c^2 - a^2) = \frac{b}{a} (a^2 + c^2 - b^2)$
$\Rightarrow a^2 (b^2 + c^2 - a^2) = b^2 (a^2 + c^2 - b^2)$
$\Rightarrow a^2b^2 + a^2c^2 - a^4 = a^2b^2 + b^2c^2 - b^4$
$\Rightarrow a^4 - b^4 + b^2c^2 - a^2c^2 = 0$
$\Rightarrow (a^2 - b^2)(a^2 + b^2) - c^2(a^2 - b^2) = 0$
$\Rightarrow (a^2 - b^2)(a^2 + b^2 - c^2) = 0$
Since $a \neq b$,we have $a^2 - b^2 \neq 0$.
Therefore,$a^2 + b^2 - c^2 = 0$,which implies $a^2 + b^2 = c^2$.
Thus,$\triangle ABC$ is a right-angled triangle.

(D) Given sides are $a=3, b=5, c=3$.
Using the Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.
Substituting the values: $\cos A = \frac{5^2 + 3^2 - 3^2}{2 \times 5 \times 3}$.
$\cos A = \frac{25 + 9 - 9}{30} = \frac{25}{30}$.
$\cos A = \frac{5}{6}$.

(B) We know that $\cos 2A = 1 - 2\sin^2 A$ and $\cos 2B = 1 - 2\sin^2 B$.
Substituting these into the expression:
$\frac{1 - 2\sin^2 A}{a^2} - \frac{1 - 2\sin^2 B}{b^2} = \frac{1}{a^2} - \frac{2\sin^2 A}{a^2} - \left(\frac{1}{b^2} - \frac{2\sin^2 B}{b^2}\right)$
$= \frac{1}{a^2} - \frac{1}{b^2} - 2 \left( \frac{\sin^2 A}{a^2} - \frac{\sin^2 B}{b^2} \right)$
By the Sine Rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = k$,so $\frac{\sin^2 A}{a^2} = \frac{\sin^2 B}{b^2} = k^2$.
Thus,$\frac{\sin^2 A}{a^2} - \frac{\sin^2 B}{b^2} = 0$.
Therefore,the expression simplifies to $\frac{1}{a^2} - \frac{1}{b^2}$.

(A) Given $\angle A = 60^{\circ}$. By the Cosine Rule,$a^2 = b^2 + c^2 - 2bc \cos A$.
Since $\cos 60^{\circ} = \frac{1}{2}$,we have $a^2 = b^2 + c^2 - 2bc \left(\frac{1}{2}\right) = b^2 + c^2 - bc$.
Now,consider the expression $(a+b+c)(b+c-a)$.
This can be rewritten as $((b+c)+a)((b+c)-a) = (b+c)^2 - a^2$.
Expanding this,we get $b^2 + c^2 + 2bc - a^2$.
Substituting $a^2 = b^2 + c^2 - bc$,we get:
$b^2 + c^2 + 2bc - (b^2 + c^2 - bc) = b^2 + c^2 + 2bc - b^2 - c^2 + bc = 3bc$.

(C) Given in $\triangle ABC$: $a=3, b=4$ and $\sin A = \frac{3}{4}$.
Using the Sine Rule: $\frac{\sin A}{a} = \frac{\sin B}{b}$.
Substituting the values: $\frac{\frac{3}{4}}{3} = \frac{\sin B}{4}$.
$\Rightarrow \frac{3}{4 \times 3} = \frac{\sin B}{4}$.
$\Rightarrow \frac{1}{4} = \frac{\sin B}{4}$.
$\Rightarrow \sin B = 1$.
Since $\sin B = 1$, we have $B = 90^{\circ}$.
Therefore, $\angle CBA = 90^{\circ}$.

(C) Given in $\triangle ABC$,$A=75^{\circ}$ and $B=45^{\circ}$.
Using the angle sum property of a triangle,$C = 180^{\circ} - (75^{\circ} + 45^{\circ}) = 60^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$.
Thus,$b = k \sin 45^{\circ} = k \frac{1}{\sqrt{2}}$ and $c = k \sin 60^{\circ} = k \frac{\sqrt{3}}{2}$.
Also,$a = k \sin 75^{\circ} = k \sin(45^{\circ}+30^{\circ}) = k \left( \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \right) = k \frac{\sqrt{3}+1}{2\sqrt{2}}$.
We need to find $b + c\sqrt{2}$.
$b + c\sqrt{2} = k \frac{1}{\sqrt{2}} + k \frac{\sqrt{3}}{2} \cdot \sqrt{2} = k \left( \frac{1}{\sqrt{2}} + \frac{\sqrt{3}}{\sqrt{2}} \right) = k \frac{\sqrt{3}+1}{\sqrt{2}}$.
Since $a = k \frac{\sqrt{3}+1}{2\sqrt{2}}$,we have $k = \frac{2\sqrt{2}a}{\sqrt{3}+1}$.
Substituting $k$ back,$b + c\sqrt{2} = \left( \frac{2\sqrt{2}a}{\sqrt{3}+1} \right) \left( \frac{\sqrt{3}+1}{\sqrt{2}} \right) = 2a$.

(D) Given: $a=6 \text{ cm}$,$b=10 \text{ cm}$,$c=14 \text{ cm}$.
Using the Law of Cosines:
$\cos C = \frac{a^2 + b^2 - c^2}{2ab} = \frac{6^2 + 10^2 - 14^2}{2 \times 6 \times 10} = \frac{36 + 100 - 196}{120} = \frac{-60}{120} = -\frac{1}{2}$.
Since $\cos C = -\frac{1}{2}$,$C = 120^{\circ}$,which is an obtuse angle.
The sum of all angles in a triangle is $180^{\circ}$.
Therefore,the sum of the remaining two angles (which must be acute) is $A + B = 180^{\circ} - 120^{\circ} = 60^{\circ}$.

(A) Let the sides of the triangle be $a = x^2+x+1$,$b = 2x+1$,and $c = x^2-1$.
Since $x^2+x+1$ is the greatest side,the greatest angle $A$ is opposite to side $a$.
Using the Law of Cosines: $\cos A = \frac{b^2+c^2-a^2}{2bc}$.
Substituting the values: $\cos A = \frac{(2x+1)^2 + (x^2-1)^2 - (x^2+x+1)^2}{2(2x+1)(x^2-1)}$.
Expanding the terms: $(2x+1)^2 = 4x^2+4x+1$,$(x^2-1)^2 = x^4-2x^2+1$,and $(x^2+x+1)^2 = x^4+x^2+1+2x^3+2x^2+2x = x^4+2x^3+3x^2+2x+1$.
Numerator: $(4x^2+4x+1) + (x^4-2x^2+1) - (x^4+2x^3+3x^2+2x+1) = -2x^3-x^2+2x+1$.
Factoring the numerator: $-(2x^3+x^2-2x-1) = -[x^2(2x+1) - 1(2x+1)] = -(x^2-1)(2x+1)$.
Thus,$\cos A = \frac{-(x^2-1)(2x+1)}{2(2x+1)(x^2-1)} = -\frac{1}{2}$.
Therefore,$A = 120^{\circ}$.

(C) Given,$b = 2$,$c = 3$,and $\angle B = \frac{\pi}{6}$.
Using the cosine rule in $\triangle ABC$:
$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$
Substituting the given values:
$\cos \frac{\pi}{6} = \frac{a^2 + 3^2 - 2^2}{2 \cdot a \cdot 3}$
$\frac{\sqrt{3}}{2} = \frac{a^2 + 9 - 4}{6a}$
$\frac{\sqrt{3}}{2} = \frac{a^2 + 5}{6a}$
Multiplying both sides by $6a$:
$3\sqrt{3} a = a^2 + 5$
Rearranging the terms,we get:
$a^2 - 3\sqrt{3} a + 5 = 0$

(C) Using the Law of Cosines,$\cos C = \frac{a^2+b^2-c^2}{2ab}$.
Given $\angle C = 60^{\circ}$,we have $\cos 60^{\circ} = \frac{1}{2} = \frac{a^2+b^2-c^2}{2ab}$.
This implies $ab = a^2+b^2-c^2$,or $a^2+b^2 = ab+c^2$.
Now,substitute $a^2+b^2$ in the given expression:
$\frac{c(a+b)+(a^2+b^2)}{(b+c)(c+a)} = \frac{ca+cb+ab+c^2}{bc+ab+c^2+ac}$.
Since the numerator and denominator are identical,the value is $1$.

(C) We are given $\Sigma(q+r) \cos P = (q+r) \cos P + (r+p) \cos Q + (p+q) \cos R$.
Expanding this,we get:
$(q \cos P + r \cos P) + (r \cos Q + p \cos Q) + (p \cos R + q \cos R)$
Rearranging the terms:
$(q \cos P + p \cos Q) + (r \cos P + p \cos R) + (r \cos Q + q \cos R)$
Using the projection law,we know that $r = q \cos P + p \cos Q$,$q = r \cos P + p \cos R$,and $p = r \cos Q + q \cos R$.
Substituting these into the expression:
$r + q + p = p + q + r$
Since the semi-perimeter $s = \frac{p+q+r}{2}$,we have $p+q+r = 2s$.
Therefore,$\Sigma(q+r) \cos P = 2s$.